Elliptic Integrals
R. I. Badran
Mathematical Physics
Complete elliptic integrals
When  

, the elliptic integrals are called the complete
2
elliptic integrals of first and second kinds.


K (k )  F (k , ) 
2

E (k )  E (k , ) 
2
2

1  k 2 sin 2 
0

d
2

1  k 2 sin 2  d
0
[Note: These integrals have special tables which are more
accurate than the tables of F(k, ) and E(k, )].
Other properties of Legendre forms of elliptic integrals:
Question: How can we find integrals with various values of ?
Answer: We must always integrate over a number of  (not
/2) intervals and then add or subtract the correct
integral over an interval of length less than /2.
1. Using the definition of complete elliptic integrals, we can
show that
F (k , n   )  2nK  F (k ,  )
E(k , n   )  2nE  E(k , )
Elliptic Integrals
R. I. Badran
Mathematical Physics
Proof: This can be proved by having a close insight into the
shown diagram:
f(sin2)
/4
7/4
2
9/4

The elliptic integrals are both functions of sin2, namely
f(sin2). If we plot this function such that  between 0 and /2
and that of  between /2 and  will give the same value. Thus
for  between 0 and  is one period of f(sin2). The rest of the
graph repeats itself.
Remember that the area under the curve

f (sin 2  )d could
be either F(k, ) or E((k, ). Now we can find the area under the
curve for  between 0 and 9/4 as follows:
9

4
2

0


2

 areaA 
0

0
4


2
4

 4 

0
0
0
Also the area under the curve for  between 0 and 7/4 as
follows:
7

0
4
2



2
 areaA 
0


0
4


2
4

4 

0
0
0
[Warning: You have to be very careful here not to confuse this
7
later area with

0
4
3


0
2


4

0
].
Elliptic Integrals
R. I. Badran
Mathematical Physics
2. The elliptic integrals can be shown to be odd functions of ,
namely:
F (k , )   F (k , )
E(k , )   E(k , )
3.
When the lower limit in the elliptic integrals is different
from zero, they can be expressed as follows:
2

1
2


1
d
1  k 2 sin 2 
d
1  k sin 
2
2

2

0
1
d
1  k 2 sin 2 

0
d
1  k 2 sin 2 
 F (k , 2 )  F (k ,1 )
Similarly
2

1
1  k 2 sin 2  d  E (k ,  2 )  E (k , 1 )
Exercise 1: Evaluate I 


0
9  8x 2
(1  x )
2 3
dx ,
[Hint: take E(1/3)=1.525].
Solution: Put x = tan  and proceed.

Exercise 2: Evaluate I 
6

0
d
1  4 sin 
2
,
[Hint: take K(1/2)=1.688].
Solution: Put 4 sin2 = sin2 and proceed…
Elliptic Integrals
R. I. Badran
Mathematical Physics
Example: The problem of simple pendulum for large angles.
Solution:
 2
Recalling the DE of motion ( ) 
2g
cos   C .

By taking a swing of any amplitude, say , rather than /2 such
that
   0 at  = , we get C  
 (  ) 2 
2g
cos  .

2g
(cos  cos  ) .

It must be noted that the period of a swing from - to  and
back is T. Thus the limits can be summarized as :
  0    

T

t

0

t


4
The above DE can be rewritten as:


0
T
d
cos   cos 

4

0
2g
2 g T
dt 

 4
To obtain a final result for T we should solve problem 17,
section 12, chap.11.

We need to evaluate the integral I  
0
Put
sin

2
 sin

2
sin 
d
.
cos  cos 
and change the limit of the integral
such that for  = 0   = 0 and for  =    = /2. We have
Elliptic Integrals
R. I. Badran

I
1
(
2
But
and
cos

2
sin 2
2
2 sin

2
cos
0
 1  sin 2

 sin 2
2

0

sin
2

2
2

2
 sin
2
 ,
2
sin 2 
sin 2   sin
1  sin 2
I  2 F (sin
 T  4
2
d
)

2
cos  .
d
2
I 2

cos 
Mathematical Physics

2
 
sin 2 
,

, )  2 K (sin )
2 2
2


K (sin )
g
2
Special cases:
i) For  not too large (i.e. < /2  sin2 /2 < ½)
[You have to solve problem 1, section 12, chap. 11].
You might get a good approximation for T when
you use the binomial expansion:
Elliptic Integrals
R. I. Badran
(1  x)n  1  nx 


n(n  1) 2 n(n  1)(n  2) 3
x 
x 
2!
3!
2

2
sin 2  )
Thus F (k , )   (1  sin
2
2
0


Mathematical Physics
1
2
d can be written as:
1

3

F (k , )   (1  sin 2 sin 2   sin 4 sin 4     )d
2
2
2
8
2
0
2

 
1 1
 3 3

K (sin ){ (1   sin 2   sin 4  )}
g
2 2
2 2
2 8 8
2
 T  4
ii) For small enough   sin /2  /2, hence

2
T  2
(1 
   )
g
16
iii)
For very small  we will reach to the
approximate result, that is,
T  2

g
x
Exercise: Evaluate the integral
I 
0
[Hint: you may reach a step with
k
1
2
].
10  5 x 2
dx .
1  x2
I  10 E (k , ) , where